Question

An ideal gas expands from 2 L to 6 L at a constant temperature of 300 K. Calculate the work done by the gas if pressure remains constant at 2 atm. (1 atm = $1.013 \times 10^5$ Pa)

• A. 810.4
• B. 820
• C. 820.5
• D. 795.5

Hint:

For constant pressure expansion, use $W = P \Delta V$; always convert atm to Pa and L to m\textsuperscript{3} before calculation.

Answer 1

The Correct Option is A

Given: \[ P = 2 \, \text{atm} = 2 \times 1.013 \times 10^5 \, \text{Pa} = 2.026 \times 10^5 \, \text{Pa} \] \[ \Delta V = V_f - V_i = 6\, \text{L} - 2\, \text{L} = 4\, \text{L} = 4 \times 10^{-3} \, \text{m}^3 \] Work done by the gas at constant pressure is: \[ W = P \Delta V = 2.026 \times 10^5 \times 4 \times 10^{-3} = 810.4 \, \text{J} \] But this doesn’t match any options — let's recheck: Actually, \[ W = 2 \, \text{atm} \times (6 - 2) \, \text{L} = 2 \times 4 = 8 \, \text{L atm} \] \[ 1\, \text{L atm} = 101.3 \, \text{J} \Rightarrow W = 8 \times 101.3 = 810.4 \, \text{J} \]

Answer 2

To solve the problem, we need to calculate the work done by an ideal gas expanding from 2 L to 6 L at constant pressure of 2 atm.

1. Formula for Work Done at Constant Pressure: 
The work done $W$ by a gas during expansion or compression at constant pressure is given by:
$ W = P \Delta V = P (V_f - V_i) $

2. Given Data:
Initial volume, $V_i = 2\, L = 2 \times 10^{-3} \, m^3$ (since $1\,L = 10^{-3} \, m^3$)
Final volume, $V_f = 6\, L = 6 \times 10^{-3} \, m^3$
Pressure, $P = 2\, atm = 2 \times 1.013 \times 10^5\, Pa = 2.026 \times 10^5\, Pa$

3. Calculating Change in Volume:
$ \Delta V = V_f - V_i = (6 - 2) \times 10^{-3} = 4 \times 10^{-3} \, m^3 $

4. Calculating Work Done:
$ W = P \times \Delta V = 2.026 \times 10^5 \times 4 \times 10^{-3} = 810.4 \, J $

Final Answer:
The work done by the gas is $ {810.4\, J} $.